How do I figure out the mean and standard deviation for the following distributi

Question

How do I figure out the mean and standard deviation for the following distribution. A survey or American Society of Investors found 30% of individual investors have used a discount broker. In a random sample of 9 individuals: what is the mean and standard deviation? How do I figure out the mean and standard deviation for the following distribution. A survey or American Society of Investors found 30% of individual investors have used a discount broker. In a random sample of 9 individuals: what is the mean and standard deviation? A survey or American Society of Investors found 30% of individual investors have used a discount broker. In a random sample of 9 individuals: what is the mean and standard deviation?

Explanation / Answer

Here we are given that,

proportion of American Society of Investors (p) = 30% = 0.30

n = sample size = 9

The normal distribution is a good approximation to the binomial when n is sufficiency large and p is not too close to 0 or 1. How large n needs to be depends on the value of p. If p is near 0.5, the approximation can be good for n much less than 20. However, it is better to be conservative and limit the use of the normal distribution as an approximation to the binomial when np > 5 and n(1 - p) > 5.

Here distribution of p is normal with mean is p and and standard deviation is sqrt((pq)/n).

mean = 0.30

sd = sqrt ((0.3*0.7) / 20) = 0.1528

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